# Is the norm of the acceleration the same as the componentwise calculation of the acceleration? [closed]

I am using the following question as an example to base my conceptual question on:

4.00-kg object has a velocity of 3.00i m/s at one instant. Eight seconds later, its velocity has increased to (8.00i + 10.0j) m/s. Assuming the object was subject to a constant total force, find (a) the components of the force and (b) its magnitude.

I thought that part b could be answered as follows. vi=3.00j t=8 sec norm(vf)=12.806

vf=vi+at 12.806=3+8a so a=1.2257 F=m*a equals 4.88 then. However, the answer is 5.59N.

Therefore, from what I see, the componentwise calculation of the acceleration is different than if the norm of the vector is taken at the beginning as I did. Is this true? If so, why?

• I can't follow your solution, but the acceleration is 1.397 not 1.225 Commented Nov 2, 2022 at 16:40
• The two velocities don't have the same direction. Adding or subtracting their magnitudes is not meaningful for acceleration.
– nasu
Commented Nov 2, 2022 at 16:58
• Please format mathematical expressions using MathJax, as it makes it much easier to read. Commented Nov 2, 2022 at 17:36

You cannot use the equation $$v_f = v_i + a.t$$ by substituting the magnitudes of $$\vec{\bf{v_f}}$$ and $$\vec{\bf{v_i}}$$, because the equations are derived from the definitions of velocity and acceleration, which are both vectors: $$\vec{\bf{a}} = \frac{\text{d}\vec{\bf{v}}}{\text{dt}} \rightarrow \text{d}\vec{\bf{v}} = \vec{\bf{a}}.\text{dt}$$ $$∴ \Delta \vec{\bf{v}} = \int_{i}^{f} \vec{\bf{a}}.\text{dt}$$ Because the force is given to be constant, the net acceleration is constant too, and hence:
$$\Delta \vec{\bf{v}} = \vec{\bf{a}}.\Delta{t} \rightarrow \boxed{\vec{\bf{v_f}} = \vec{\bf{v_i}} + \vec{\bf{a}}.\Delta{t}}$$
Consequently from this equation, you may write: $$|\vec{\bf{v_f}} - \vec{\bf{v_i}}| = |\vec{\bf{a}}|.\Delta{t}$$ But not $$\color{red}{|\vec{\bf{v_f}}| = |\vec{\bf{v_i}}| + |\vec{\bf{a}}|.\Delta{t}}$$